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The equipartition theorem provides a convenient way to derive a relation between the hamiltonian of an ideal gas to the temperature of the system. For an extreme relativistic ideal gas, the kinetic energy of a single particle is given by the formula

$$H=c\sqrt{p_x^2+p_y^2+p_z^2}.$$

Then one can write the average

$$\langle H\rangle=\langle p_x\frac{\partial H}{\partial p_x}\rangle+\langle p_y\frac{\partial H}{\partial p_y}\rangle+\langle p_z\frac{\partial H}{\partial p_z}\rangle= 3 k_{\small\text{B}} T$$

where the last equality follows from the equipartition formula.

Is it possible to get a similar relation for the second moment of the energy $\langle H^2\rangle$?

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1 Answer 1

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Similar to the formula $$ \langle p_x \frac{\partial H}{\partial p_x} \rangle = k_BT, $$ there exists the following relation $$ \langle p_x \frac{\partial H^2}{\partial p_x} \rangle = 2 k_BT \langle H \rangle + 2(k_B T)^2. \quad (1) $$ For the hamiltonian of an ultrarelativistic particle, we have $$ H^2 = \frac12\left(p_x \frac{\partial H^2}{\partial p_x} + p_y \frac{\partial H^2}{\partial p_y} + p_z \frac{\partial H^2}{\partial p_z} \right) $$ Together with $\langle H \rangle = 3k_BT$, this gives $$ \langle H^2 \rangle = 12(k_BT)^2. $$

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  • $\begingroup$ Thank you. Do you have a reference where I can look up the second identity? $\endgroup$
    – user56224
    Commented Apr 14, 2020 at 7:51
  • $\begingroup$ @kaffeeauf, I don't have a reference. I've derived it myself. Do you know how to derive a relation for $<p_x\partial H/\partial p_x>$? The same method gives a relation for $H^2$. $\endgroup$
    – Gec
    Commented Apr 14, 2020 at 8:03
  • $\begingroup$ The formula (1) is general. It is applicable to any reasonable hamiltonian $H$. $\endgroup$
    – Gec
    Commented Apr 14, 2020 at 8:13

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